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Chapter

Fifteen

McGraw-Hill/Irwin

© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

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Chapter FifteenNonparametric Methods: Chi-Square Nonparametric Methods: Chi-Square ApplicationsApplications

GOALSWhen you have completed this chapter, you will be able to:ONE

List the characteristics of the Chi-square distribution.

TWO Conduct a test of hypothesis comparing an observed set of frequencies to an expected set of frequencies.

THREEConduct a hypothesis test to determine whether two classification criteria are related.

Goals

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Characteristics of the Chi-Square Distribution

The major characteristics of the chi-square distribution are:

It is positively skewedIt is non-negativeThere is a family of chi-square distributions

Chi-Square ApplicationsChi-Square Applications

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df = 3

df = 5

df = 10

2 distribution

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Goodness-of-Fit Test: Equal Expected Frequencies

Let f0 and fe be the observed and expected frequencies respectively.

H0: There is no difference between the observed and expected frequencies.

H1: There is a difference between the observed and the expected frequencies.

e

eo

f

ff 22The test statistic is:

The critical value is a chi-square value with (k-1) degrees of freedom, where k is the number of categories

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Example 1 continued

Day of Week Number Absent

Monday 120

Tuesday 45

Wednesday 60

Thursday 90

Friday 130

Total 445

The following information shows the number of employees absent by day of the week at a large a manufacturing plant. At the .01 level of significance, is there a difference in the absence rate by day of the week?

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H0: There is no difference between the observed and expected frequencies.

H1: There is a difference between the observed and the expected frequencies

This is given in the problem as .01.

It is the chi-square distribution.

Example 1 continued

Step 1Step 1: State the null and alternate hypotheses

Step 2Step 2: Select the level of significance.

Step 3Step 3: Select the test statistic.

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EXAMPLE 1 continued

Assume equal expected frequency as given in the problem

fe = (120+45+60+90+130)/5=89

The degrees of freedom: (5-1)=4

The critical value of 2 is 13.28. Reject the null and accept the alternate ifComputed 2 > 13.28

or p< .01

Step 4Step 4: Formulate the decision rule.

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Example 1 continued

Day Frequency Expected (fo – fe)2/fe

Monday 120 89 10.80

Tuesday 45 89 21.75

Wednesday 60 89 9.45

Thursday 90 89 0.01

Friday 130 89 18.89

Total 445 445 60.90

Step Five:Step Five: Compute the value of chi-square and make a decision.

The p(2 > 60.9) = .000000000001877 or essentially 0.

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Example 1 continued

We conclude that there is a difference in the number of workers absent by day of the week.

Because the computed value of chi-square, 60.90, is greater than the critical value, 13.28, the p of .000000000001877 < .01, H0 is rejected.

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Example 2

The U.S. Bureau of the Census indicated that 63.9% of the population is married, 7.7% widowed, 6.9% divorced (and not re-married), and 21.5% single (never been married). A sample of 500 adults from the Philadelphia area showed that 310 were married, 40 widowed, 30 divorced, and 120 single. At the .02 significance level can we conclude that the Philadelphia area is different from the U.S. as a whole?

Goodness-of-fit TestGoodness-of-fit Test: Unequal Expected Frequencies

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Example 2 continued

Step 4: H0 is rejected if 2 >9.837, df=3, or if

p of .02

Step 1: H0: The distribution has not changed H1: The distribution has changed.

Step 2: The significance level given is .02.

Step 3: The test statistic is the chi-square.

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Calculate the expected

frequencies

Married: (.639)500 = 319.5Widowed: (.077)500 = 38.5Divorced: (.069)500 = 34.5Single: (.215)500 = 107.5

Example 2 continued

Status

Married 310 319.5 .2825

Widowed 40 38.5 .0584

Divorced 30 34.5 .5870

Single 120 107.5 1.4535

Total 500 2.3814

Calculate chi-square

values.

f0 fe ( ) /f f fe e02

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Step 5: 2 = 2.3814, p(2 > 2.3814) = .497.

Example 2 continued

The null hypothesis is not rejected. The distribution regarding marital status in

Philadelphia is not different from the rest of the United States.

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Contingency Table Analysis

A contingency contingency tabletable is used to investigate whether two traits or characteristics are related.

Chi-square can be used to test for a relationship between two nominal scaled variables, where one variable is independent of the other.

Contingency Table AnalysisContingency Table Analysis

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Each observation is classified according to two criteria.We use the usual hypothesis testing procedure.The degrees of freedomdegrees of freedom are equal to:

(number of rows-1)(number of columns-1).The expected frequencyexpected frequency is computed as:

Expected Frequency = (row total)(column total)

grand total

Contingency Table AnalysisContingency Table Analysis

Contingency table analysis

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Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the .05 level of significance, can we conclude that gender and the location of the accident are related? Example 3

Contingency Table AnalysisContingency Table Analysis

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Step 4: The degrees of freedom equal (r-1)(c-1) or 2. The critical 2 at 2 d.f. is 9.21. If computed 2 >9.21, or if p < .01, reject the null and accept the alternate.

Step 5: A data table and the following contingency table are constructed.

Step 1: H0: Gender and location are not related.

H1: Gender and location are related.

Step 2: The level of significance is set at .01.

Step 3: the test statistic is the chi-square distribution.

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Example 3 continued

The expected frequency for the work-male intersection is computed as (90)(80)/150=48.Similarly, you can compute the expected frequencies for the other cells.

Observed frequencies (fo )

Gender Work Home Other Total

Male 60 20 10 90

Female 20 30 10 60

Total 80 50 20 150

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Expected frequencies (fe )

Gender Work Home Other Total

Male (80)(90)150

= 48

(50)(90)

150

=30

(20)(90)

150

=12

90

Female (80)(60)

150

=32

(50)(60)

150

=20

(20)(60)

150

=8

60

Total 80 50 20 150

Example 3 continued

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2: (fo – fe)2/ fe

Gender Work Home Other Total 2

Male (60-48)2

48

(20-30)2

30

(10-12)2

12

6.667

Female (20-32)2

32

(30-20)2

20

(12-10)2

10

10.000

Total 16.667

Example 3 continued

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The p(2 > 16.667) = .00024.

Since the 2 of 16.667 > 9.21, p of .00024 < .01, reject the null and conclude that there is a relationship between the location of an accident and the gender of the person involved.

Example 3 concluded